Question

A sample of 300ml of a solution, magnesium chloraye Mg(ClO3)2, has a molar concentration if 4.02 mol/L. How much mass (grams) of magnesium coloration is in this sample?

Answer 1

To find the mass of magnesium chlorate (Mg(ClO₃)₂) in the solution, we can use the formula:

\[
\text{mass} = \text{molarity} \times \text{volume} \times \text{molar mass}
\]

1. **Molarity (C)**: Given as 4.02 mol/L.
2. **Volume (V)**: Given as 300 mL. We convert this to liters:

\[
300 \text{ mL} = 0.300 \text{ L}
\]

3. **Molar Mass (MM)** of Mg(ClO₃)₂:
- Magnesium (Mg): 24.305 g/mol
- Chlorine (Cl): 35.453 g/mol
- Oxygen (O): 16.00 g/mol

Calculating molar mass:
- For Mg(ClO₃)₂: 1 Mg + 2 Cl + 6 O
\[
\text{MM} = 1(24.305) + 2(35.453) + 6(16.00)
\]
\[
= 24.305 + 70.906 + 96.00
\]
\[
= 191.211 \text{ g/mol}
\]

4. **Calculate the mass:**
\[
\text{mass} = (4.02 \text{ mol/L}) \times (0.300 \text{ L}) \times (191.211 \text{ g/mol})
\]
\[
= 4.02 \times 0.300 \times 191.211
\]
\[
= 4.02 \times 57.3633
\]
\[
= 230.47 \text{ g}
\]

Thus, the mass of magnesium chlorate in the sample is approximately **230.47 grams.**